A rigid tank contains 20 lbm of air at 20 psia and 70°F. More air is added to the tank until the pressure and temperature rise to 23.5 psia and 90°F, respectively. Determine the amount of air added to the tank. The gas constant of air is R

Answer:

The new rms speed is 2 times the original rms speed.

Explanation:

The root‑mean‑square speed, rms, is related to temperature, , by the formula

[tex]_{rms}[/tex]= √3 / ℳ

For a given gas,  

[tex]_{rms}[/tex] ∝ √

or

[tex]_{rms,2}[/tex] / [tex]_{rms,1}[/tex] = √[tex]_{2}[/tex] / [tex]_{1}[/tex]

In this case, is quadrupled.

√4 = 2

The new rms speed is 2 times the original rms speed.

The root-mean-square (rms) speed of gas molecules is proportional to the square root of the temperature. If the absolute temperature is quadrupled, the new rms speed is 2 times the original rms speed.

Effect of Temperature on Root Mean Square Speed of Gas Molecules

The root-mean-square (rms) speed, [tex]V_{rms[/tex] , of the molecules in a gas is related to the absolute temperature, T, by the equation:

[tex]V_{rms[/tex] = [tex]\sqrt((3kBT) / m)[/tex]

where kB is the Boltzmann constant and m is the mass of a molecule. This shows that Vrms is proportional to the square root of the temperature. If the temperature T is quadrupled, the new temperature T' is 4T.

Therefore, the new rms speed [tex]V_{rms[/tex]' becomes:

[tex]V_{rms[/tex]' = [tex]\sqrt((3kB * 4T) / m)[/tex]= [tex]\sqrt(4) * \sqrt((3kBT) / m)[/tex] = 2 * [tex]V_{rms[/tex]

This means that the new rms speed is 2 times the original rms speed.

Answer:

-56.4 kJ/mol

Explanation:

There are two heat flows in this experiment.

Heat released by reaction + heat absorbed by solution = 0

                     q1                   +                     q2                     = 0

                   nΔH                 +                  mCΔT                  = 0

1. Moles of water formed

                  KOHL + H2SO4 → K2SO4 + 2H2O

V/mL:          40.0        40.0

c/(mol/L)      1.00       0.500

   Moles of KOH = 40.0 mL × (1.00 mmoL/1 mL)     = 40.00 mmol KOH

Moles of H2SO4 = 40.0 mL × (0.500 mmoL/1 mL) = 20.00 mmol H2SO4

Moles of H2O from KOH

= 40.00 mL KOH × (2.00 mmol H2O/2 mmoL KOH)  = 40.00 mmol H2O

Moles H2O from of H2SO4

= 20.00 mL × (2 mmoL H2O/1 mmol H2SO4)             = 40.00 mmol H2O

KOH and H2SO4 are present in equimolar amounts.

They form 40.00 mmol = 0.040 00 mol of water.

2. Calculate q1

q1 = 0.04000 mol × ΔH = 0.040 00ΔH J

3. Calculate q2

Mass of water formed = 0.040 00 mol × (18.02 g/1 mol) = 0.7208 g

                  V(solution) = 40.00 + 40.00 = 80.00 mL

        Mass of solution = 80.00 mL × (1.02 g/1 mL) = 81.60 g

                 Total mass = 81.60 + 0.7208 = 82.32 g

                              ΔT = T2 - T1 = 27.85 - 21.00 = 6.85 °C

                              q2 = 82.32 × 4.00 × 6.85 = 2256 J

4. Calculate ΔH

0.040 00ΔH + 2256 = 0

            0.040 00ΔH = -2256

                           ΔH = -2256/0.040 00 = = -56 400 J/mol = -56.4 kJ/mol

The enthalpy of neutralization is -56.4 kJ/mol.

Answer:

Joule-Thomson coefficient for an ideal gas:

[tex]\mu_{J.T} = 0[/tex]

Explanation:

Joule-Thomson coefficient can be defined as change of temperature with respect to pressure at constant enthalpy.

Thus,

[tex]\mu_{J.T} = \left [\frac{\partial T}{\partial P} \right ]_H[/tex]

Also,

[tex]H= H (T,P)[/tex]

[tex]Differentiating\ it,[/tex]

[tex]dH= \left [\frac{\partial H}{\partial T}\right ]_P dT + \left [\frac{\partial H}{\partial P}\right ]_T dT[/tex]

Also, [tex] C_p[/tex] is defined  as:

[tex]C_p = \left [\frac{\partial H}{\partial T}\right ]_P[/tex]

[tex]So,[/tex]

[tex]dH= C_p dT + \left [\frac{\partial H}{\partial P}\right ]_T dT[/tex]

Acoording to defination, the ethalpy is constant which means [tex]dH = 0[/tex]

[tex]So,[/tex]

[tex]\left [\frac{\partial H}{\partial P}\right ]_T = -C_p\times \left [\frac{\partial T}{\partial P}\right ]_H[/tex]

[tex]Also,[/tex]

[tex]\mu_{J.T} = \left [\frac{\partial T}{\partial P}\right ]_H[/tex]

[tex]So,[/tex]

[tex]\left [\frac{\partial H}{\partial P}\right ]_T =-\mu_{J.T}\times C_p[/tex]

For an ideal gas,

[tex]\left [\frac{\partial H}{\partial P}\right ]_T = 0[/tex]

So,

[tex]0 =-\mu_{J.T}\times C_p[/tex]

Thus, [tex]C_p[/tex] ≠0. So,

[tex] \mu_{J.T} = 0[/tex]

The Joule-Thompson Coefficient is zero for an ideal gas because the temperature of the gas doesn't change during an adiabatic expansion or compression due to the lack of intermolecular forces, which is a characteristic as per the ideal gas laws.

The Joule-Thompson Coefficient measures the change in temperature of a gas when it is forced to expand or contract at constant enthalpy. In an ideal gas, the internal energy, which includes information about temperature, is only a function of temperature. Since an ideal gas follows the ideal gas law, when a gas expands or contracts with no heat exchange (an adiabatic process), its temperature remains constant due to an absence of intermolecular forces. This is because any work done on (or by) the gas doesn’t result in a change in temperature.

We know from the equation of internal energy (Eint = 3nRAT/2), that any change in internal energy is due to a change in temperature (ΔT), and if ΔT = 0 for an isothermal process, then the change in internal energy of the system (ΔEint) also equals zero, indicating no change in temperature. Hence, for an ideal gas, the Joule-Thompson Coefficient is zero because the temperature of the gas doesn't change during an adiabatic expansion or compression.

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Answer:

Rate = 1.09*10^-3 (mol H2/L)/s

Explanation:

Given:

Initial concentration of H2, C1 = 0 M

Final concentration of H2, C2 = 0.101 M

Time taken, t = 93.0 s

To determine:

The rate of the given reaction

Calculation:

The decomposition of PH3 is represented by the following chemical reaction

[tex]4 PH3(g)\rightarrow P4(g) + 6 H2(g)[/tex]

Reaction rate in terms of the appearance of H2 is given as:

[tex]Rate = +\frac{1}{6}*\frac{\Delta [H2]]}{\Delta t}[/tex]

[tex]Rate = +\frac{1}{6}*\frac{C2[H2]-C1[H2]}{\Delta t}[/tex]

Here C1(H2) = 0 M and C2(H2) = 0.101 M

Δt = 93.0 s

[tex]Rate = \frac{(0.101-0.0)M}{93.0 s} =1.09*10^{-3} M/s[/tex]

Since molarity M = mole/L

rate = 1.09*10^-3 (mol H2/L)/s

Answer: The correct answer is [tex]4molH_2\times \frac{2molH_2O}{2molH_2}=4 molH_2O[/tex]

Explanation:

We are given:

Moles of hydrogen gas = 4 moles

As, oxygen is given in excess. Thus, is considered as an excess reagent and hydrogen is considered as a limiting reagent because it limits the formation of products.

For the given chemical equation:

[tex]2H_2(g)+O_2(g)\rightarrow 2H_2O(l)[/tex]

By Stoichiometry of the reaction:

2 moles of hydrogen produces 2 moles of water molecule.

So, 4 moles of hydrogen will produce = [tex]\frac{2molH_2O}{2molH_2}\times 4molH_2=4mol[/tex] of water.

Hence, the correct answer is [tex]4molH_2\times \frac{2molH_2O}{2molH_2}=4 molH_2O[/tex]

Hey there!:

Aqueous ethanol contains water , sodium borohydride will react with water and decompose to give hydrogen gas :

NHB4 + 2 H2O => NaBO2 + 4 H2

thus a large excess of borohydride should be used used to take into account losses from the reaction with water , so that there is sufficient sodium borohydride left for the reduction of the chemical compound os interest .

Hope this helps!

The importance of using a large excess of sodium borohydride in a reduction in aqueous ethanol lies in its role as a reducing agent that donates hydrogen ions. However, in an aqueous solution, some of the sodium borohydride may react with water in a hydrolysis reaction, forming hydrogen gas and borate ions. Using an excess ensures an ample amount for the intended reaction.

It's important to use a large excess of sodium borohydride when doing a reduction in aqueous ethanol because sodium borohydride (NaBH₄) is an excellent reducing agent. It donates hydrogen ions (H⁻) which do not survive in an acidic medium. The H⁻ ions react initially, and then the acid is added to donates a proton to oxygen (O) in the second step.

However, the aqueous environment complicates this process because water can react with sodium borohydride in a hydrolysis reaction. Water acts as an incoming nucleophile, and in the presence of water, NaBH4 may end up reducing water molecules, forming hydrogen gas (H₂) and borate ions (BO3⁻). Using an excess of sodium borohydride ensures that there's enough of it to react with the intended substance, even when a portion of it reacts with water.

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Answer: 2800 g

Explanation:

[tex]CaCO_3(s)\rightarrow CaO(s)+CO_2(g)[/tex]

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Given mass = 5 kg = 5000 g

[tex]\text{Number of moles}=\frac{5000g}{100g/mol}=50moles[/tex]

1 mole of [tex]CaCO_3[/tex] produces = 1 mole of [tex]CaO[/tex]

50 moles of [tex]CaCO_3[/tex] produces =[tex]\frac{1}{1}\times 50=50moles[/tex] of [tex]CaO[/tex]

Mass of [tex]CaO=moles\times {\text{Molar mass}}=50moles\times 56g/mole=2800g[/tex]

2800 g of [tex]CaO[/tex] is produced from 5.0 kg of limestone.

A decomposition reaction splits the reactants into two or more products. The mass of the quicklime produced from 5 kg of limestone is 2800 gm.

Mass is the amount or the weight of the substance occupied in the system. It can be calculated in grams or kilograms.

The decomposition reaction of the Limestone can be shown as:

[tex]\rm CaCO_{3} \rightarrow CaO + CO_{2}[/tex]

The number of the mole of limestone is given as:

[tex]\rm Moles = \rm \dfrac {Mass}{Molar \;mass}[/tex]

Here, mass is 5000 gm and the molar mass is 100 g/mol

Substituting values in the equation above:

[tex]\begin{aligned}\rm n &= \dfrac{5000}{100}\\\\&= 50\;\rm mol\end{aligned}[/tex]

The stoichiometry coefficient of the reaction gives:

1 mole of limestone = 1 mole quicklime

So, 50 moles of limestone = x moles of quicklime

Solving for x:

[tex]\begin{aligned}\rm x &= \dfrac{50 \times 1}{1}\\\\&= 50 \;\rm mole\end{aligned}[/tex]

Mass of quicklime is calculated as:

[tex]\begin{aligned}\rm mass &= \rm moles \times \rm molar \; mass\\\\&= 50 \times 56\\\\&= 2800\;\rm gm\end{aligned}[/tex]

Therefore, 2800 gm of quicklime is produced.

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Answer : The time taken for the reaction is, 28 s.

Explanation :

Expression for rate law for first order kinetics is given by :

[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]

where,

k = rate constant  = 0.0632

t = time taken for the process  = ?

[tex][A_o][/tex] = initial amount or concentration of the reactant  = 1.28 M

[tex][A][/tex] = amount or concentration left time 't' = [tex]1.28\times \frac{17}{100}=0.2176M[/tex]

Now put all the given values in above equation, we get:

[tex]0.0632=\frac{2.303}{t}\log\frac{1.28}{0.2176}[/tex]

[tex]t=28s[/tex]

Therefore, the time taken for the reaction is, 28 s.

The equilibrium pressure of H2 is 0.96 atm and the impossible solution of the quadratic equation is -1.379.

The equilibrium pressure of H2 is calculated by creating ICE table as follows;

            2 N H3 ( g ) ⟷ N2( g ) + 3H2

I:           1                         1              1

C:         -2x                      x             3x

E:        1 - 2x                    1 + x         1 + 3x

[tex]KP = \frac{(N_2)(H_2)^3}{(NH_3)^2} \\\\0.83 = \frac{(1 + x)(1 + 3x)^3}{(1 - 2x)^2}[/tex]

0.83(1 - 2x)² = (1 + x)(1 + 3x)³

0.83(1 - 4x + 4x²) = (1 + x)((1 + 3x)³)

0.83 - 3.32x + 3.32x² = (1 + x)((1 + 3x)³)

0.83 - 3.32x + 3.32x² = 1 + 10x + 36x² + 54x³ + 27x⁴

27x⁴ + 54x³ + 32.68x² + 13.32x + 0.17 = 0

x = -1.379 or - 0.013

H2 = 1 + 3(-1.379)

H2 = -3.13 atm

H2 = 1 + 3(-0.013)

H2 = 0.96 atm

Thus, the equilibrium pressure of H2 is 0.96 atm and the impossible solution of the quadratic equation is -1.379.

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Answer:

For a: The molality and molarity of the given solution is 2.45m and 2.28 M respectively.

For b: The molarity of the solution when more water is added is 0.912 M

Explanation:

To calculate the molality of solution, we use the equation:

[tex]Molarity=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}[/tex]

Where,

[tex]m_{solute}[/tex] = Given mass of solute [tex](C_3H_8O_3)[/tex] = 42.0 g

[tex]M_{solute}[/tex] = Molar mass of solute [tex](C_3H_8O_3)[/tex] = 92.093 g/mol

[tex]W_{solvent}[/tex] = Mass of solvent (water) = 186 g

Putting values in above equation, we get:

[tex]\text{Molality of }C_3H_8O_3=\frac{42\times 1000}{92.093\times 186}\\\\\text{Molality of }C_3H_8O_3=2.45m[/tex]

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]    .....(1)

We are given:

Molarity of solution = ?

Molar mass of [tex](C_3H_8O_3)[/tex] = 92.093 g/mol

Volume of solution = 200 mL

Mass of [tex](C_3H_8O_3)[/tex] = 42 g

Putting values in above equation, we get:

[tex]\text{Molality of }C_3H_8O_3=\frac{42\times 1000}{92.093\times 200}\\\\\text{Molality of }C_3H_8O_3=2.28M[/tex]

Hence, the molality and molarity of the given solution is 2.45m and 2.28 M respectively.

Now, the 300 mL water is added to the solution. So, the total volume of the solution becomes (200 + 300) = 500 mL

Using equation 1 to calculate the molarity of solution, we get:

Molar mass of [tex](C_3H_8O_3)[/tex] = 92.093 g/mol

Volume of solution = 500 mL

Mass of [tex](C_3H_8O_3)[/tex] = 42 g

Putting values in equation 1, we get:

[tex]\text{Molality of }C_3H_8O_3=\frac{42\times 1000}{92.093\times 500}\\\\\text{Molality of }C_3H_8O_3=0.912M[/tex]

Hence, the molarity of the solution when more water is added is 0.912 M

Answer:

4600 Liters NH₃(g)

Explanation:

Applying Avogadro's law to the reaction of hydrogen and nitrogen gas forming ammonia, we find that 6.9 m3 of hydrogen will form about 4.6 m3 of ammonia, assuming that temperature and pressure remain constant.

The subject in question relates to the application of Avogadro's law to chemical reactions involving gases. More specifically, we're investigating the reaction between hydrogen gas and nitrogen gas to produce ammonia gas. As per Avogadro's law, gases react in definite and simple proportions by volume, if all gas volumes are measured at the same temperature and pressure.

Focusing on the hydrogen to ammonia conversion, the reaction equation N₂(g) + 3H₂(g) turns into 2NH3(g). We can surmise that three volumes of hydrogen gas (H2) react to form two volumes of ammonia gas (NH3). Considering that 6.9 m3 of hydrogen gas is consumed, by the rule of three, we can infer that the reaction would result in approximately 4.6 m3 of ammonia gas.

Bear in mind, these calculations are assuming that the temperature and pressure remain constant during the reaction and that it goes to completion.

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Explanation:

According to Faraday's law, the amount of a substance deposited or liberated in electrolysis process is proportional to the quantity of electric charge passed and to the equivalent weight of the substance.

Formula to calculate the mass of substance liberated according to Faraday's law is as follows.

             m = [tex](\frac{Q}{F})(\frac{M}{Z})[/tex]

where,          m = mass of substance liberated at electrode

                     Q = electric charge passing through the substance

                     F = Faraday constant = 96,487 C [tex]mol^{-1}[/tex]

                     M = molar mass of the substance

                     Z = valency number of ions of the substance

Since, it is given that mass is 3 kg or 3000 g (as 1 kg = 1000 g), molar mass of Al is 27, Z is 3.

Therefore, putting the values in the above formula as follows.

                      m = [tex](\frac{Q}{F})(\frac{M}{Z})[/tex]

                     3000 g = [tex](\frac{Q}{96,487 C mol^{-1}})(\frac{27}{3})[/tex]

                        Q = 32162333.33 C

As it is given that V = 4.50 Volt. Also, it is known that

                       Energy = [tex]V \times Q[/tex]

Therefore, calculate the energy as follows.

                      Energy = [tex]V \times Q[/tex]

                                   = [tex]4.50 V \times 32162333.33 C[/tex]

                                   = 144730500 J

As it is known that [tex]3.6 \times 10^{6}[/tex] J = 1 KW Hr

So, convert 144730500 J into KW Hr as follows.

                   [tex]\frac{144730500 J \times 1 KW Hr}{3600000 J}[/tex]

                          = 40.202 KW Hr

Thus, we can conclude that the number of kilowatt-hours of electricity required to produce 3.00 kg of aluminum from electrolysis of compounds from bauxite is 40.202 KW Hr when the applied emf is 4.50V.

The production of aluminum from bauxite involves several chemical reactions and an electrolysis process in a Hall-Héroult cell. Aluminum oxide is reduced to aluminum metal during electrolysis. The exact amount of electricity required cannot be stated without additional industrial data.

The process of making aluminum from bauxite involves several stages of chemical reactions. Initially, bauxite, AlO(OH), reacts with hot sodium hydroxide to form soluble sodium aluminate, leaving behind impurities. Aluminum hydroxide is then precipitated out and heated to form aluminum oxide, Al2O3, which is dissolved in a molten mixture of cryolite and calcium fluoride. An electrolytic cell, specifically called the Hall-Héroult cell, is used for the electrolysis process. During electrolysis, the reduction of aluminum ions to aluminum metal takes place at the cathode, while oxygen, carbon monoxide, and carbon dioxide form at the anode. The amount of electricity required to produce 3 kg of aluminum using a 4.5V emf cannot be directly calculated without additional data such as the exact chemical conversion efficiencies and energy losses in the process, which are typically proprietary information for aluminum production companies.

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Answer: The volume of liquid in liters is 2.5812 L.

Explanation:

Density of an object is defined as the ratio of its mass and volume. The chemical equation representing density of an object is:

[tex]\text{Density of an object}=\frac{\text{Mass of an object}}{\text{Volume of an object}}[/tex]

We are given:

Mass of liquid = 3.02 kg = 3020 g     (Conversion factor: 1kg = 1000 g)

Density of liquid = [tex]1.17 g/cm^3[/tex]

Putting values in above equation, we get:

[tex]1.17g/cm^3=\frac{3020g}{\text{Volume of liquid}}\\\\\text{Volume of liquid}=2581.2cm^3[/tex]

Converting this into liters, we use the conversion factor:

[tex]1L=1000cm^3[/tex]

So, [tex]\Rightarrow \frac{1L}{1000cm^3}\times 2581.2cm^3[/tex]

[tex]\Rightarrow 2.5812L[/tex]

Hence, the volume of the liquid is 2.5812 L.

The volume in liters of 3.02 kg of the liquid is 2.58 Liters

Density is the ratio of mass to volume of a substance. The density of a substance is given by:

Density = mass / volume

Given that a liquid has a density of 1.17 g/cm³ and a mass of 3.02 kg, the volume is:

Density = 1.17 g/cm³ = 1.17 kg/L

Density = mass/volume

1.17 = 3.02/volume

Volume = 3.02/1.17

Volume = 2.58 Liters

Hence the volume in liters of 3.02 kg of the liquid is 2.58 Liters

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Answer:

The annual production of carbon dioxide is [tex]9.12\times 10^{10} kg[/tex].

Explanation:

Distance covered by each car = 5790 miles

Rate of consumption of gasoline =24. mile/gal

For every 24.1 mile 1 gallon of gasoline is used

Gasoline used by a single car by travelling 5790 miles =

[tex]\frac{1}{24.1}\times 5790 mile=240.24 gal[/tex]

Number of cars in the United states = 40.0 million = [tex]4\times 10^7[/tex]

Total gallons of gasoline consumed by 40 million cars = [tex]4\times 10^7\times 240.24 gal=9.60\times 10^9 gal[/tex]

1 gallon of gasoline produces = 9.50 kg of [tex]CO_2[/tex]

Then [tex]9.60\times 10^9 gal[/tex] of gasoline will produce:

[tex]9.60\times 10^9\times 9.50 kg =9.12\times 10^{10} kg[/tex] of [tex]CO_2[/tex]

The annual production of carbon dioxide is [tex]9.12\times 10^{10} kg[/tex].

Answer:

Explanation:

1) Data:

a) Tb₁ = 100.000°C

b) Kb = 0.512 °C/m

c) mass of solute = 13.62 g

d) mass of solvent = 217.5 g

e) Tb₂ = 100.094°C

f) Solute: nonvolatile and nonelectrolyte

g) MM = ?

2) Chemical principles and formulae:

a) The boiling point elevation of a non-volatile solute is a colligative property, which follows this equation:

Where:

b) Molality, m:

c) Molar mass, MM:

3) Solution:

i) ΔTb = Tb₂ - Tb₁ = 100.094°C - 100.000°C = 0.094°C

ii) ΔTb = Kb × m ⇒ m = ΔTb / Kb = 0.094°C / (0.512°C/m) = 0.1836 m

iii) m = number of moles of solute / kg of solvent ⇒

    number of moles of solute = m × kg of solvent = 0.1836 m × 0.2175 kg

    number of moles of solute = 0.03993 mol

iv) MM = mass in grams / number of moles = 13.62 g / 0.03993 mol = 341.1 g/mol

Final answer:

The molecular weight of the synthesized compound is calculated using the change in boiling point and the molal boiling point elevation constant to first determine molality and then the number of moles of the compound. The molecular weight is then found by dividing the mass of the compound by the moles of the compound, yielding a value of 341.1 g/mol.

Explanation:

To calculate the molecular weight of the synthesized compound, we apply the boiling point elevation formula ΔT = Kb × m, where ΔT is the change in boiling point, Kb is the molal boiling point elevation constant of water, and m is the molality of the solution. Given Kb for water is 0.512°C/m and ΔT is 0.094°C (100.094°C - 100.000°C), we can calculate the molality:

m = ΔT / Kb = 0.094°C / 0.512°C/m = 0.1836 m

Next, we calculate the moles of solute using molality and the mass of the solvent (water):

moles of solute = molality × mass of solvent in kg = 0.1836 m × 0.2175 kg = 0.0399 mol

The molecular weight (MW) is then found by dividing the mass of the compound by the moles of the compound:

MW = mass of compound / moles of compound = 13.62 g / 0.0399 mol = 341.1 g/mol

Answer: c. oxidation - gain of electrons

Explanation:

1. Synthesis reaction is a chemical reaction in which two reactants are combining to form one product.

Example: [tex]Li_2O+CO_2\rightarrow Li_2CO_3[/tex]  

2. Decomposition is a type of chemical reaction in which one reactant gives two or more than two products.

Example: [tex]Li_2CO_3\rightarrow Li_2O+CO_2[/tex]

3. Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.

[tex]M\rightarrow M^{n+}+ne^-[/tex]

4. Dehydration reaction is defined as the reaction in which water is lost as product.

[tex]CH_3CH_2OH\rightarrow CH_2=CH_2+H_2O[/tex]

5. Hydrolysis  reaction is defined as the reaction in which water is used for decomposition.

Example: [tex]CH_3COOCH_2CH_3+H_2O\rightarrow CH_3COOH+CH_3CH_2OH[/tex]

The mismatched pair in the list is 'oxidation - gain of electrons.' Oxidation is actually characterized by the loss of electrons, not the gain. All other pairs accurately depict the respective chemical reactions.

The question asked is trying to identify which of the provided pairs misrepresents a type of chemical reaction. For the majority of these reactions, the descriptions are accurate:

However, the provided definition of an oxidation reaction is incorrect. Oxidation is characterized by the loss of electrons, not the gain. The oxidation-reduction (redox) reactions involve an exchange of electrons between reactants, where the substance losing electrons is being oxidized, and the substance gaining electrons is being reduced.

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Answer:

[tex]H_2PO_3^-[/tex] is Bronsted Lowry base.

[tex]H_2O[/tex] is Bronsted Lowry acid.

Explanation:

According to the Bronsted Lowry conjugate acid-base theory:

[tex]H_2PO_3^-(aq)+H2O(l)\rightarrow H_3PO_3(aq)+OH^-(aq)[/tex]

[tex]H_2PO_3^-[/tex] is Bronsted Lowry base.It accepts protons and forms conjugate acid [tex]H_3PO_3[/tex]

[tex]H_2O[/tex] is Bronsted Lowry acid.It donates protons and forms conjugate base [tex]OH^-[/tex]

[tex](CH_3)_2NH(g)+BF_3(g)\rightarrow (CH_3)_2NHBF_3(s)[/tex]

There in no exchange of proton in an above reaction.Neither of the reactants and products are Bronsted Lowry acid or Bronsted Lowry base

Answer:

CuSO4 + Fe -> FeSO4 + Cu

Explanation:

This reaction is a classic example of a redox reaction. I won't go in too deep, but the basic thing is that electrons from the Fe atom go to the Cu2+ ion. Therefore, Fe becomes an ion, and Cu - an electroneutral atom:

Fe + Cu2+ -> Fe2+ + Cu.

Silver is not a very reactive metal and it does not give up its electrons to Cu.

Iron displaces copper in Copper(II) sulfate solution, forming Iron sulfate and Copper. The reaction with the silver rod does not occur as silver is less reactive than copper.

When the student placed an iron nail into the copper(II) sulfate solution, a displacement reaction occurred. Iron is higher on the reactivity series than copper, so it displaced copper from the copper(II) sulfate. The balanced chemical equation for this reaction is: Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s)

On the other hand, when the silver rod was placed into the copper(II) sulfate solution, no visible reaction occurred. This is because Silver is less reactive than Copper and cannot displace it from its sulfate salt.

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Answer:

For a: The amount of heat transferred for the given amount of methanol is 94.6736 kJ.

For b: The mass of methane gas produced will be 10.384 g.

Explanation:

For the given chemical reaction:

[tex]2CH_3OH(g)\rightarrow 2CH_4(g)+O_2(g);\Delta H=+252.8kJ[/tex]

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of methanol = 24.0 g

Molar mass of methanol = 32.04 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of methanol}=\frac{24.0g}{32.04g/mol}=0.749mol[/tex]

By Stoichiometry of the reaction:

For every 2 moles of methanol, the amount of heat transferred is +252.8 kJ.

So, for every 0.749 moles of methanol, the amount of heat transferred will be = [tex]\frac{252.8}{2}\times 0.749=94.6736kJ[/tex]

Hence, the amount of heat transferred for the given amount of methanol is 94.6736 kJ.

By Stoichiometry of the reaction:

252.8 kJ of energy is absorbed when 2 moles of methane gas is produced.

So, 82.1 kJ of energy will be absorbed when = [tex]\frac{2}{252.8}\times 82.1=0.649mol[/tex] of methane gas is produced.

Now, calculating the mass of methane gas from equation 1, we get:

Molar mass of methane gas = 16 g/mol

Moles of methane gas = 0.649 moles

Putting values in equation 1, we get:

[tex]0.649mol=\frac{\text{Mass of methane gas}}{16g/mol}\\\\\text{Mass of methane}=10.384g[/tex]

Hence, the mass of methane gas produced will be 10.384 g.

To calculate the amount of heat transferred, use the equation q = mcΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. To calculate the grams of methane gas produced, use the equation q = ΔH, where ΔH is the enthalpy change during the reaction.

To calculate the amount of heat transferred, we need to use the equation q = mcΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. In this case, we are working at constant pressure, so we can use the equation q = ΔH. First, calculate the moles of CH3OH(g) using the molar mass. Then, use the molar ratio from the balanced equation to determine the moles of CH4(g). Finally, use the molar mass of CH4(g) to calculate the mass of CH4(g). Substitute the values into the equation q = ΔH to find the amount of heat transferred.

The given enthalpy change during the reaction is +82.1 kJ. Using the equation q = ΔH, we can calculate the moles of CH4(g). Then, use the molar mass of CH4(g) to calculate the mass of CH4(g).

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Answer: The [tex]\Delta H^o_{formation}[/tex] for the reaction is 591.9 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical reaction for the formation reaction of [tex]N_2O[/tex] is:

[tex]N_2(g)+\frac{1}{2}O_2(g)\rightarrow N_2O(g)[/tex]    [tex]\Delta H^o_{formation}=?[/tex]

The intermediate balanced chemical reaction are:

(1) [tex]2NH_3(g)+3N_2O(g)\rightarrow 4N_2(g)+3H_2O(l)[/tex]    [tex]\Delta H_1=-1010kJ[/tex]    ( ÷  3)

(2) [tex]4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(l)    [tex]\Delta H_2=1531kJ[/tex] ( ÷  6)

Reversing Equation 1 and then adding both the equations, we get the enthalpy change for the chemical reaction.

[tex]\Delta H^o_{formation}=[\frac{\Delta H_1}{3}]+[\frac{\Delta H_2}{6}][/tex]

Putting values in above equation, we get:

[tex]\Delta H^o_{formation}=[\frac{1010}{3}]+[\frac{1531}{6}]\\\\\Delta H^o_{formation}=591.9kJ[/tex]

Hence, the [tex]\Delta H^o_{formation}[/tex] for the reaction is 591.9 kJ.

Hess's law is defined as the sum of amount of heat absorbed or released in the given chemical equation remains constant, irrespective of the steps involved in the reaction. The [tex]\Delta \text H^0_{\text{formation}}&=591.9 \text{kJ}[/tex] is for the given reaction.

Given that,

Now, the given chemical equations are:

The intermediate reactions between the above equation are:

Reversing the equation and then adding both the values, the enthalpy change becomes:

Therefore, the enthalpy change of the reaction is [tex]\Delta \text H^0_{\text{formation}}&=591.9 \text{kJ}[/tex].

To know more about enthalpy change, refer to the following link:

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Answer : The normal boiling point of ethanol will be, [tex]348.67K[/tex] or [tex]75.67^oC[/tex]

Explanation :

The Clausius- Clapeyron equation is :

[tex]\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})[/tex]

where,

[tex]P_1[/tex] = vapor pressure of ethanol at [tex]30^oC[/tex] = 98.5 mmHg

[tex]P_2[/tex] = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg

[tex]T_1[/tex] = temperature of ethanol = [tex]30^oC=273+30=303K[/tex]

[tex]T_2[/tex] = normal boiling point of ethanol = ?

[tex]\Delta H_{vap}[/tex] = heat of vaporization = 39.3 kJ/mole = 39300 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

[tex]\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})[/tex]

[tex]T_2=348.67K=348.67-273=75.67^oC[/tex]

Hence, the normal boiling point of ethanol will be, [tex]348.67K[/tex] or [tex]75.67^oC[/tex]

Answer: 28.3 g/mol

Explanation:

According to the ideal gas equation:'

[tex]PV=nRT[/tex]

P= Pressure of the gas = 0.951 atm

V= Volume of the gas = 280 mL = 0.28 L     (1L=1000 ml)

T= Temperature of the gas = 25°C=(25+273)K=298 K   (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm/K mol

[tex]n=\frac{PV}{RT}=\frac{0.951\times 0.28L}{0.0821 \times 298}=0.010moles[/tex]

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]

[tex]0.010=\frac{0.283}{\text {Molar mass}}[/tex]

[tex]{\text {Molar mass}}=28.3g/mol[/tex]

Thus the molar mass of the gas is 28.3 g/mol.

The molar mass of the gas is 25.73 g/mol. This was calculated using the ideal gas law and conversion of units, finally using the formula molar mass = mass in g/moles.

To calculate the molar mass of the gas, we can use the ideal gas law, which is PV = nRT. Here, n = number of moles, R = ideal gas constant, T = temperature in Kelvin, P = pressure and V = volume. First, convert the temperature from Celsius to Kelvin by adding 273.15: 25.0°C + 273.15 = 298.15 K. We need to find n, the number of moles. From the ideal gas law, we know that n = PV/RT. Substituting the given values: n = (0.951 atm * 280 mL) / (0.0821 L atm mol⁻¹K⁻¹ * 298.15 K) = 0.011 mol.

The mass given is 0.283 g. So, using the relation molar mass = mass in g/moles; molar mass of gas = 0.283g / 0.011 mol = 25.73 g/mol.

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Hey there!:

In Kjeldahl's method  estimation of amount of nitrogen in a sample, the sample is first digested by using sulphuric acid , which converts the nitrogen in the sample to ammonium sulphate.

The ammonium ions in the digested product are converted to ammonia gas by adding excess of base like NaOH. The ammonia gas thus produced is collected by condensation.

The amount of ammonia produced is estimated by titration with standard solution of acid. We are using 0.1 N HCl in this case.

Volume of 0.1 N HCl used = 11.95 mL

Volume of of 0.1 N HCl used in blank titration =0.1 mL

Therefore,actual volume us of 0.1 N HCl

used by analyte( ammonia solution) = 11.95 mL - 0.1 mL  

= 11.85 mL

0.1 N HCl = 0.1 M HCl , as HCl is a monobasic acid and therefore its molar mass will be equal to its equivalent mass.

Therefore, concentration of HCl solution used = 0.1 M = 0.1 mol L-1

Volume of HCl solution used =11.85 mL =0.01185 L

Nº.of moles of  HCl used =  0.1 mol L-1  * 0.01185 L

= 0.001185 moles

The equation for reaction between NH3 (aq) and HCl during titration is  :

NH3 (aq) + HCl (aq) <= >  NH4Cl

From the above equation, we see that each mole of HCl reacts completely with 1 mole of NH3 during titration.

Therefore, no. of moles of NH3 that would have been neutralized by 0.001185 moles of HCl = 0.001185 moles

From the formula of  NH3 , 1 mole of  NH3 contains 1 mole of N atoms.

Therefore, nº of moles of :

N - atoms present = 0.001185 moles

Moar mass of N= 14 g/mol

Mass of 0.001185 moles of N =0.001185 moles * 14 g/ mol

=  0.01659 g

% of N in protein is given as 16%.

16%.of mass of protein  = mass of nitrogen in protein sample  = 0.01659 g

(16/100) *mass of protein  =  0.01659 g

mass of protein  =  0.1036875 g

% of protein  in potato flour = [mass of protein in  potato flour / mass of  potato flour] * 100%

( 0.1036875 g / 5.724 ) * 100%

=  1.812 %

Hope this helps!

Final answer:

The percent protein was found to be 1.8113 %.

Explanation:

To calculate the percent protein in the potato flour, we first need to correct the titration volume for the blank, then figure out the amount of Nitrogen (N) in the potato flour and finally convert this to percent protein using the factor that proteins in potato flour contain 16% nitrogen. Since each gram of nitrogen corresponds to 6.25 grams of protein (since 100/16 = 6.25), we can use this conversion factor to get the final percent protein.

Volume of HCl used for the sample is 11.95 mL

Volume of HCl used for the blank is 0.1 mL

Net volume of HCl used for the sample is 11.85 mL (11.95 mL - 0.1 mL)

Since the concentration of HCl is 0.1 N, we multiply the net volume by the normality to find the milliequivalents (meq) of nitrogen:
11.85 mL * 0.1 N = 1.185 meq

To find the grams of nitrogen, we use the fact that 1 meq of N equals 0.014 g:
1.185 meq * 0.014 g/meq = 0.01659 g of N

To convert grams of N to percent nitrogen in the sample, we divide by the sample mass and multiply by 100:
(0.01659 g / 5.724 g) * 100 = 0.2898 % N

Multiply the percent N by the conversion factor to get percent protein:
 0.2898 % N * 6.25 = 1.8113 % Protein

Answer:

low supply of oxaloacetate in the citric acid cycle

Explanation:

When there is low supply of oxalo acetate the acetyl CoA gets converted to ketone bodies to enter the TCA cycle.

In the condition of low level of glucose, the supply of oxaloacetate decreases. Thus making it unavailable to react with acetyl CoA, in this condition ketogenesis occur i.e. acetyl CoA gets converted to ketone bodies.

Answer : The final volume of oxygen gas will be, 1.292 ml

Explanation :

Charles' Law : It is defined as the volume of gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

[tex]V\propto T[/tex]

or,

[tex]\frac{V_1}{V_2}=\frac{T_1}{T_2}[/tex]

where,

[tex]V_1[/tex] = initial volume of oxygen gas = 1.75 ml

[tex]V_2[/tex] = final volume of oxygen gas = ?

[tex]T_1[/tex] = initial temperature of gas = [tex]97.8^oC=273+97.8=370.8K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]0.7^oC=273+0.7=273.7K[/tex]

Now put all the given values in the above formula, we get the final volume of the oxygen gas.

[tex]\frac{1.75ml}{V_2}=\frac{370.8K}{273.7K}[/tex]

[tex]V_2=1.292ml[/tex]

Therefore, the final volume of oxygen gas will be, 1.292 ml

To find the equilibrium concentration of Cl2 in the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g), we use the equilibrium constant and initial concentrations to set up and solve a quadratic equation. The resulting equilibrium concentration of Cl2 is 0.054 M.

To calculate the equilibrium concentration of Cl2 from the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g), let's denote the change in concentration of Cl2 as 'x' when the system reaches equilibrium. Since Cl2 and PCl3 start at 0.0571 M and PCl5 starts at 0 M, the equilibrium concentrations will be 0.0571 - x for PCl3 and Cl2, and x for PCl5.

The equilibrium constant expression is Kc = [PCl5]/[PCl3][Cl2]. Plugging in the equilibrium concentrations and the given Kc value of 0.021, we get:
0.021 = x / (0.0571 - x)2. After solving this quadratic equation, we find that the equilibrium concentration of Cl2 rounded to the nearest thousandth is 0.054 M.

Answer:

It will take 6.6 hours for 75% of the lead to decay.

Explanation:

The radioactive decay follows first order rate law

The half life and rate constant are related as

[tex]k=rate constant=\frac{0.693}{halflife}=\frac{0.693}{3.3}=0.21h^{-1}[/tex]

The rate law for first order reaction is

[tex]time=\frac{1}{k}(ln[\frac{A_{0}}{A_{t}}][/tex]

Where

A0 = initial concentration = 1 g

At= final concentration = 0.25 g (as 75% undergoes decay so 25% left]

[tex]time=\frac{1}{0.21}(ln(\frac{1}{0.25})=6.6hours[/tex]

Answer:

27.77 kJ/mol is the activation energy for the reaction.

Explanation:

Rate of the reaction at 323 K =[tex]k_1=k[/tex]

Rate of the reaction at 373 K =[tex]k_2=4k[/tex]

Activation energy for the reaction is calculated by formula:

[tex]\log \frac{k_2}{k_1}=\frac{E_a}{2.303\times R}[\frac{T_2-T_1}{T_2\times T_1}][/tex]

[tex]E_a[/tex] = Activation energy

[tex]T_1[/tex] = Temperature when rate of the reaction was [tex]k_1[/tex]

[tex]T_2[/tex] = Temperature when rate of the reaction was [tex]k_2[/tex]

Substituting the values:

[tex]\log \frac{4k}{k}=\frac{E_a}{2.303\times 8.314 J /mol K}[\frac{373 K-323K}{373 K\times 323 K}][/tex]

[tex]E_a=27,776.98 J/mol=27.77 kJ/mol[/tex]

27.77 kJ/mol is the activation energy for the reaction.

Final answer:

Without specific numeric values for the individual atomic heat capacities of the elements in sodium carbonate, directly answering the heat capacity multiple-choice question using Kopp's rule is not feasible. Additional data is required to estimate the heat capacity of solid sodium carbonate.

Explanation:

The question regarding the heat capacity of solid sodium carbonate (Na2CO3) can be answered by applying Kopp's rule, which is a method used to estimate the heat capacities of solids. According to Kopp's rule, the heat capacity of a compound in the solid state is the sum of the atomic heat capacities of the individual elements that compose the compound. Atomic heat capacities can be estimated using the Law of Dulong and Petit, which suggests that at room temperature, the molar heat capacity of many solid elements roughly equates to 3R, where R is the gas constant with a value approximately equal to 8.314 J/(mol·K).

Applying Kopp's rule would involve calculating the total molar heat capacity for sodium, carbon, and oxygen in sodium carbonate, based on their atomic weights and expected contributions. However, since specific numeric values are not provided in the question or reference information to apply Kopp's rule directly, directly answering the multiple-choice question is not feasible without additional data on the individual atomic heat capacities of the elements involved in sodium carbonate.